Let \(f:(X,\mathcal T)\to(Y,\mathcal T')\) be a continuous map. A section of \(f\) is a continuous map \(\sigma:(Y,\mathcal T')\to(X,\mathcal T)\) so that \(f\circ\sigma=\operatorname{Id}_Y\). Show that if \(f\) has a section, then \(f\) is a quotient map.
Solution
1. \(f\) is surjective.
For every \(y\in Y\), \(f(\sigma(y))=y\).
2. If \(V\subseteq Y\) and \(f^{-1}(V)\in\mathcal T\), then \(V\in\mathcal T'\).
For \(y\in Y\), \(y\in\sigma^{-1}(f^{-1}(V))\iff f(\sigma(y))\in V\iff y\in V\), because \(f\circ\sigma=\operatorname{Id}_Y\). So \(V=\sigma^{-1}(f^{-1}(V))\), which is open because \(\sigma\) is continuous.
3. Q.E.D.
\(f\) is continuous by hypothesis, surjective by step 1, and by step 2 a subset of \(Y\) is open whenever its preimage is open; so \(f\) is a quotient map.